Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

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The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

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Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.math.*;

public class Main {

    static Scanner input = new Scanner(new BufferedReader(new InputStreamReader(System.in)));

    public static void main(String[] args) {
        int n = input.nextInt();
        int m = input.nextInt();
        int[][] field = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                field[i][j] = input.nextInt();
            }
        }
        int minInRow = 500;
        for (int i = 0; i < m; i++) {
            if (field[0][i] < minInRow) {
                minInRow = field[0][i];
            }
        }
        int minMoves = 100000000;
        ans lastAns = new ans(-1);
        for (int i = minInRow; i >= 0; i--) {
            ans res = rowsAndCol(field, i);
            if (minMoves > res.res && res.res != -1) {
                minMoves = res.res;
                lastAns = res;
            }
        }
        if (minMoves == 100000000) {
            System.out.println(-1);
        } else {
            System.out.println(minMoves);
            for (int i = 0; i < lastAns.rowMoves.length; i++) {
                for (int j = 0; j < lastAns.rowMoves[i]; j++) {
                    System.out.println("row " + (i + 1));
                }
            }
            for (int i = 0; i < lastAns.colMoves.length; i++) {
                for (int j = 0; j < lastAns.colMoves[i]; j++) {
                    System.out.println("col " + (i + 1));
                }
            }
        }
    }

    static ans rowsAndCol(int[][] field, int k) {
        int[] colMoves = new int[field[0].length];
        int[] rowMoves = new int[field.length];
        int sum = k;
        rowMoves[0] = k;
        for (int i = 0; i < colMoves.length; i++) {
            if (field[0][i] < k) {
                return new ans(-1);
            }
            colMoves[i] = field[0][i] - k;
            sum += field[0][i] - k;
        }
        for (int i = 1; i < rowMoves.length; i++) {
            if (field[i][0] < colMoves[0]) {
                return new ans(-1);
            }
            rowMoves[i] = field[i][0] - colMoves[0];
            sum += field[i][0] - colMoves[0];
        }
        for (int i = 0; i < rowMoves.length; i++) {
            for (int j = 0; j < colMoves.length; j++) {
                if (field[i][j] != colMoves[j] + rowMoves[i]) {
                    return new ans(-1);
                }
            }
        }
        ans tmp = new ans(sum);
        tmp.colMoves = colMoves;
        tmp.rowMoves = rowMoves;
        return tmp;
    }

    static class ans {

        int res;
        int[] colMoves;
        int[] rowMoves;

        ans(int k) {
            res = k;
        }
    }
}